A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of a

Q: Why does a closed pipe only have odd harmonics?

Because the closed end must be a displacement node (no air movement) and the open end must be an antinode (maximum movement). This boundary condition restricts the allowed wavelengths to $\lambda/4, 3\lambda/4, 5\lambda/4$, etc., which correspond to odd multiples of the fundamental frequency.

Q: What is the difference between harmonics and overtones?

Harmonics are integer multiples of the fundamental frequency. Overtones are the actual higher resonant frequencies present in the system. In a closed pipe, the 1st overtone is actually the 3rd harmonic ($3f_0$).

Q: Why did we use $v/4L$ instead of $v/2L$?

The formula $v/2L$ is used for systems with identical boundary conditions at both ends, like a pipe open at both ends or a string fixed at both ends. A closed pipe has mixed boundary conditions, leading to the $v/4L$ formula.

Q: Does the diameter of the pipe affect the frequency?

In reality, yes. The antinode forms slightly outside the open end, requiring an 'end correction' to the length $L$. However, in standard textbook problems where the diameter is not given, we assume the pipe is narrow enough to ignore this correction.

Q: What if the calculated frequency was exactly 1250 Hz?

The question specifically asks for frequencies that lie 'below 1250 Hz'. This implies a strict inequality ($f < 1250$). If a harmonic landed exactly on 1250 Hz, it would not be counted.

Question

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Options:

(A) 12

(B) 8

(C) 6

(D) 4

Sigma Percentile · Accepted Answer

### Visualizing the Closed Pipe

Imagine an organ pipe that is closed at one end. When air vibrates inside it, a standing wave is formed. The closed end strictly restricts air movement, creating a displacement **node**, while the open end allows maximum freedom, creating an **antinode**. This physical constraint is the key to understanding the frequencies the pipe can produce.

### The Master Equation

For a closed pipe, the natural frequencies (or harmonics) are strictly **odd multiples** of the fundamental frequency. The general formula is given by:

$$f_n = \frac{(2n+1)v}{4L}$$

where $n = 0, 1, 2, \dots$, $v$ is the speed of sound, and $L$ is the length of the pipe. Let's find the fundamental frequency ($f_0$), which is the lowest possible frequency. We substitute the given values: the speed of sound $v = 340 	ext{ m/s}$, and the length $L = 0.85 	ext{ m}$.

$$f_0 = \frac{340}{4 	imes 0.85}$$

Four times $0.85$ gives us $3.4$. Dividing $340$ by $3.4$, we get exactly $100 	ext{ Hz}$. So, the fundamental frequency is $100 	ext{ Hz}$.

### Counting the Harmonics

Now, remember that a closed pipe only produces odd harmonics. This means the possible frequencies will be $100 	ext{ Hz}, 300 	ext{ Hz}, 500 	ext{ Hz}, 700 	ext{ Hz}$, and so on. Notice how the first overtone is three times the fundamental.

The question asks for the number of natural oscillations whose frequencies lie strictly below $1250 	ext{ Hz}$. We need to count how many of our odd harmonics fit into this range. Let's list them out:

- $100 	ext{ Hz}$
- $300 	ext{ Hz}$
- $500 	ext{ Hz}$
- $700 	ext{ Hz}$
- $900 	ext{ Hz}$
- $1100 	ext{ Hz}$

The next harmonic would be $1300 	ext{ Hz}$, which exceeds our limit. So, we stop at $1100 	ext{ Hz}$.

### Final Conclusion

Counting them up, we have exactly **6 valid frequencies**. Therefore, there are 6 possible natural oscillations for this air column below the given frequency limit.

We found our answer, but think about this: what if the pipe was open at both ends? How would the harmonics change? An open pipe supports all harmonics, both even and odd. Try calculating the number of valid frequencies for that case!

Animated Solution for Physics - Waves: A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Select Answer:

Visualized Solution

\text{Visualizing the Closed Pipe}

\text{Fundamental Frequency Formula}

\text{Calculating Fundamental Frequency } (f_0)

\text{Computing } f_0

\text{Higher Harmonics}

\text{Applying the Constraint}

\text{Counting the Frequencies}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Visualizing the Closed Pipe

The Master Equation

Counting the Harmonics

Final Conclusion

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